Pop Operation on Stack in DSA Python - Execution Trace

Concept Flow - Pop Operation on Stack

Check if stack is empty?

No↓

Remove top element

↓

Update top pointer

↓

Return removed element

↓

Done

Yes↓

Stack Underflow - cannot pop

↓

Done

The pop operation first checks if the stack is empty. If not, it removes the top element, updates the top pointer, and returns the removed element. If empty, it signals underflow.

Execution Sample

DSA Python

stack = [10, 20, 30]
top = 2
popped = stack.pop()
top -= 1
print(stack)

This code removes the top element from the stack and updates the top pointer.

Execution Table

Step	Operation	Stack Before	Top Pointer Before	Element Removed	Top Pointer After	Stack After	Visual State
1	Check if stack is empty	[10, 20, 30]	2	N/A	2	[10, 20, 30]	Stack not empty, proceed to pop
2	Remove top element	[10, 20, 30]	2	30	2	[10, 20]	Element 30 removed from top
3	Update top pointer	[10, 20]	2	30	1	[10, 20]	Top pointer moved down to index 1
4	Return removed element	[10, 20]	1	30	1	[10, 20]	Pop operation complete, returned 30
5	Check if stack is empty (empty case)	[]	-1	N/A	-1	[]	Stack empty, cannot pop, underflow

💡 Pop stops after removing the top element and updating the top pointer, or stops immediately if stack is empty.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
stack	[10, 20, 30]	[10, 20, 30]	[10, 20]	[10, 20]	[10, 20]
top	2	2	2	1	1
popped	N/A	N/A	30	30	30

Key Moments - 3 Insights

Why do we check if the stack is empty before popping?

Because popping from an empty stack causes an error (underflow). The execution_table row 1 shows the check prevents this by stopping if empty.

Why does the top pointer decrease after popping?

Because the top pointer always points to the last element in the stack. After removing the top element (row 2), the pointer moves down by one (row 3) to reflect the new top.

What happens to the removed element after pop?

The removed element is returned by the pop operation (row 4) and no longer part of the stack (stack state changes in row 2).

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'top' after step 3?

A -1

B 2

C 1

D 0

💡 Hint

Check the 'Top Pointer After' column in row 3 of execution_table.

At which step does the stack actually remove the top element?

A Step 1

B Step 2

C Step 3

D Step 4

💡 Hint

Look at the 'Operation' and 'Element Removed' columns in execution_table.

If the stack is empty, what happens according to the execution_table?

A Underflow occurs and pop stops

B Pop removes an element

C Top pointer increases

D Stack adds a new element

💡 Hint

Refer to the last row in execution_table describing the empty stack case.

Concept Snapshot

Pop Operation on Stack:
- Check if stack is empty (underflow check)
- Remove top element
- Decrease top pointer by 1
- Return removed element
- Stack shrinks by one element

Full Transcript

The pop operation on a stack first checks if the stack is empty to avoid errors. If not empty, it removes the top element, updates the top pointer to the next element below, and returns the removed element. If the stack is empty, it signals underflow and stops. The top pointer always points to the current top element index. After popping, the stack size decreases by one.

Practice

1. What does the pop() operation do in a stack?

easy

A. Clears all elements from the stack

B. Adds a new element to the bottom of the stack

C. Returns the bottom element without removing it

D. Removes and returns the top element of the stack

Solution

1. Step 1: Understand stack behavior

   A stack follows Last In First Out (LIFO) principle, so the last added element is on top.

2. Step 2: Define pop operation

   The pop() removes and returns this top element, reducing stack size by one.

3. Final Answer:

   Removes and returns the top element of the stack -> Option D

4. Quick Check:

   pop() = remove top element [OK]

Hint: Pop always removes the last added item [OK]

Common Mistakes:

• Thinking pop removes bottom element
• Confusing pop with push operation
• Assuming pop returns the whole stack

2. Which of the following is the correct syntax to pop an element from a stack named stack in Python?

easy

A. stack.pop(0)

B. pop(stack)

C. stack.pop()

D. stack.remove()

Solution

1. Step 1: Recall Python list pop syntax

   In Python, pop() is a method called on the list object to remove the last element.

2. Step 2: Check each option

   stack.pop() uses stack.pop() which is correct. pop(stack) is invalid syntax. stack.pop(0) pops first element, not top of stack. stack.remove() is invalid method.

3. Final Answer:

   stack.pop() -> Option C

4. Quick Check:

   stack.pop() is correct syntax [OK]

Hint: Use stack.pop() to remove top element [OK]

Common Mistakes:

• Using pop with index 0 removes bottom, not top
• Calling pop as a standalone function
• Using remove() which deletes by value, not position

3. What will be the output of the following code?

stack = [10, 20, 30]
popped = stack.pop()
print(popped)
print(stack)

medium

A. 30
[10, 20]

B. 10
[20, 30]

C. 20
[10, 30]

D. 30
[20, 10]

Solution

1. Step 1: Identify initial stack and pop operation

   Stack starts as [10, 20, 30]. Calling pop() removes last element 30 and returns it.

2. Step 2: Check printed values

   Variable popped is 30. Stack after pop is [10, 20]. So print outputs 30 and then [10, 20].

3. Final Answer:

   30 [10, 20] -> Option A

4. Quick Check:

   pop() returns last element 30 [OK]

Hint: pop() returns last item, stack shrinks [OK]

Common Mistakes:

• Thinking pop removes first element
• Confusing popped value with stack after pop
• Misreading list order after pop

4. Find the error in the following code that tries to pop an element from an empty stack:

stack = []
popped = stack.pop()
print(popped)

medium

A. SyntaxError due to missing parentheses

B. IndexError because pop on empty list is invalid

C. TypeError because pop requires an argument

D. No error, prints null

Solution

1. Step 1: Understand pop on empty list

   Calling pop() on an empty list raises an IndexError because there is no element to remove.

2. Step 2: Check error types

   SyntaxError and TypeError do not apply here. The code does not print null; it raises an error.

3. Final Answer:

   IndexError because pop on empty list is invalid -> Option B

4. Quick Check:

   pop() on empty list = IndexError [OK]

Hint: Pop on empty stack causes IndexError [OK]

Common Mistakes:

• Assuming pop returns null if empty
• Confusing error type with SyntaxError
• Not checking if stack is empty before pop

5. Given a stack implemented as a list, how can you safely pop all elements without causing errors, and print them in order they were removed?

hard

A. Use a while loop: while stack: print(stack.pop())

B. Use a for loop: for i in range(len(stack)): print(stack.pop())

C. Use recursion without base case to pop elements

D. Directly print stack.pop() without checking if empty

Solution

1. Step 1: Understand safe popping

   Using while stack: checks if stack is not empty before popping, avoiding errors.

2. Step 2: Evaluate options

   Use a while loop: while stack: print(stack.pop()) safely pops all elements printing each. Use a for loop: for i in range(len(stack)): print(stack.pop()) uses fixed range which fails if stack size changes during loop. Use recursion without base case to pop elements causes infinite recursion. Directly print stack.pop() without checking if empty risks error if stack empty.

3. Final Answer:

   Use a while loop: while stack: print(stack.pop()) -> Option A

4. Quick Check:

   while stack: pop safely [OK]

Hint: Check stack non-empty before pop in loop [OK]

Common Mistakes:

• Using for loop with changing stack size
• Not checking empty stack before pop
• Infinite recursion without base case